YES

1 decompositions
 #0 -----------
   1: a() -> b()
   2: a() -> d()
   3: b() -> a()
   4: c() -> a()
   5: c() -> b()

@Rule Labeling
--- R
   1: a() -> b()
   2: a() -> d()
   3: b() -> a()
   4: c() -> a()
   5: c() -> b()

--- S
   1: a() -> b()
   2: a() -> d()
   3: b() -> a()
   4: c() -> a()
   5: c() -> b()